Find the partial fraction decomposition for each rational express...

Question

Find the partial fraction decomposition for each rational expression. See Examples 1–4. (2x^5 + 3x^4 - 3x^3 - 2x^2 + x)/(2x^2 + 5x + 2)

Accepted Answer

Hello, today we're gonna be using partial fraction decomposition to simplify the following rational expression. So what we are given is three X to the fifth minus 19 X to the fourth plus 37 X cubed minus 21 X squared plus X plus one. And all of that is going to be over three X squared minus 16 X plus 21. Now, one thing to note about the current rational expression is that the the highest value exponent in the numerator is greater than the highest value exponent in the denominator. So what this means is that we're first going to either to simplify the rational expression before we use partial fraction decomposition. Now, in order to do this, you would simplify this by using long division. And in a previous video, we have discussed on how to perform long division. So if you need a little bit of help with that topic, you can go ahead and catch that video before. But when you do perform long division on the current rational expression, you should end up with the quantity X cubed minus X squared plus X plus one over the original denominator, which is three X squared minus 16 X plus 21. Now, for now, we're gonna group up these first two terms of X cubed minus X squared. And what we want to focus on now is using partial fraction decomposition to go ahead and simplify the remainder. So currently, we have X plus one divided by three X squared minus 16 X plus 21. In order to use partial fraction decomposition, we need to first make sure that our denominator is in a fully factored form. Now, three X squared minus 16 X plus 21 is A, is in the form of a factor trinomial that is going to factor to give us the quantity three X minus seven times X minus three. And we're still going to keep the same numerator, which is X plus one. Now, what we currently have here in the denominator are two non repeating linear expressions and because they are two nonlinear expressions, we're gonna go ahead and use that to set up the partial fractions decomposition. So we're going to have the remainder which is X plus 1/3, X minus seven times X minus three. And we're gonna set up the first partial fraction for three X minus seven. Now, because that's a non repeating linear quantity that is going to give us the partial fraction of A over three X minus seven. And now we're going to set up the next partial fraction for X minus three. Now, X minus three is a non repeating linear quantity. So that is going to give us the partial fraction of B over X minus three. So currently, what we've done is we've set up this equation X plus 1/3, X minus seven times X minus three is equal to the sum of the two partial fractions. But now what we wanna do is we want to go ahead and solve for the coefficients A and B. Now, in order to do that, we're going to need to multiply this entire equation by a lowest common denominator. That way, we can simplify the equation. Now, the lowest common denominator between all the denominators is going to be the product of all the possible factors. So we have X minus three, we have three X minus seven and we have three X minus seven times X minus three. A product of all the possible combinations of the denominators is going to give us the lowest common denominator of three X minus seven times X minus three. So what we want to go ahead and do is we want to multiply every term in this equation by the lowest common denominator that is going to give us X plus 1/3 X minus seven times X minus three, multiplied by the lowest common denominator, which is going to be three X minus seven times X minus three. And for now we're gonna go ahead and put this over one that is going to equal to our first partial fraction, which is a over three X minus seven times the L CD, which is going to be three X minus seven times X minus three over one plus B over X minus three times the L CD, which is going to be three X minus seven times X minus three. So now what we're going to do is we're going to go ahead and use cross multiplication in order to cancel out the, the denominators. So in the first term, we have three X minus seven times X minus three, in both the numerator and denominator, we can go ahead and cross cancel those quantities. In the second term, we have three X minus seven in both the numerator and denominator. So we can go ahead and cross out those quantities. And in the last term, we have X minus three in both the numerator and denominator. So we can eliminate those terms. So what we're currently left with is this equation X plus one is equal to A times the quantity X minus three plus B times the quantity three X minus seven. So now what we're going to do is we're going to simplify the right hand side of this current equation. In order to do this, we're going to first distribute A into X minus three and B into three, X minus seven. That is going to give us a times X minus three, A plus three B X minus seven B. Now, here I recommend that we go ahead and organize our terms. So we're gonna group up our X terms and we're gonna group up our constants together. Reordering our terms is going to give us A X plus three B X minus three A minus seven B. And now we're gonna go ahead and factor out an X from these first two terms and group up the constants together doing this allows us to rewrite the right hand side of the equation as X times A plus three B plus negative three A minus seven B. And keep in mind that this is all still equal to X plus one. Now, here comes the tricky part about solving the coefficient of AM B. We can go ahead and use the current equation to set up a system of equations that will allow us to solve for A and B. So keep in mind that the first group has the coefficient of X. So we're going to set the coefficient of X equal to the coefficient of X on the left hand side. Now, on the left hand side, X has a coefficient of one. And on the right hand side, X has a following coefficient of A plus three B. So we can go ahead and create an equation which is going to be A plus three B is equal to one. And we're going to do the same thing to the constant. We have the constant on the right hand side, which is negative three A minus seven B. And that is going to equal to the constants on the left hand side of the equation, which in this case is positive one. So this is going to equal to one. Now, what we can go ahead and do is we can go ahead and solve this using the method of the system of equations. And so if we take a look at the 1st and 2nd equation, we have negative three A and we have positive A. If we were to multiply the first equation by three and add the two equations together, we can go ahead and eliminate the A term. So we're going to multiply the first equation by three that is going to give us three A plus nine B is equal to three. And we didn't change the second equation. So we're going to rewrite it as negative three A minus seven B is equal to one. Adding these two equations together three A minus three A is going to cancel each other out nine B minus seven B is going to give us two B and three plus one is going to give us four. Now, if we divide both sides of the equation by two, this is going to give us the value of B is equal to two. And that's gonna be the value of the first coefficient of the partial fractions. Now that we have a value for B, we can go ahead and plug this into one of the original equations to solve for A. If we plug this into the first equation, we get A plus three times B. And since B is equal to two, this is going to be three times two is equal to one, three times two is going to give us six. And if we subtract six to both sides of the equation, we get A is equal to negative five. So now that we have the coefficients of A and B, we're going to go ahead and plug this back into the original setup for our partial fractions. Now recall that the partial fractions we set up was a over three X minus seven plus B over X minus three. If we plug in the current values of A and B that we saw for, we get negative 5/3, X minus seven plus two over X minus three, this is going to be the partial fraction decomposition of the remainder. However, this is not our full solution because keep in mind that in the beginning of the problem, we had to do long division of the rational expression. So we also need to include the terms which are X X cubed minus X squared. So if we combine those two terms, the full solution to the problem is going to be X cubed minus X squared minus 5/3, X minus seven plus two over X minus three. And this is going to be the complete simplification of the original rational expression. And with that being said, the answer to this problem is going to be D so I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

Find the partial fraction decomposition for each rational expression. See Examples 1–4. (2x^5 + 3x^4 - 3x^3 - 2x^2 + x)/(2x^2 + 5x + 2)

Key Concepts

Rational Expressions

Partial Fraction Decomposition

Polynomial Long Division

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